Question

A first order reaction has a rate constant of 2.303 × 10⁻³ s⁻¹. The time required for 40 g of this reactant to reduce to 10 g will be ______.

Options

• 602 s

• 230.3 s

• 301 s

• 2000 s

Answer 1

A first order reaction has a rate constant of 2.303 × 10⁻³ s⁻¹. The time required for 40 g of this reactant to reduce to 10 g will be 602 s.

Explanation:

Given: Rate constant (k) = 2.303 × 10⁻³ s⁻¹

Initial concentration (or mass) ([R]₀) = 40 g

Final concentration (or mass) ([R]) = 10 g

First order integrated law is given as

`t = 2.303/k log  [R]_0/[R]`

⇒ `t = 2.303/(2.303 xx 10^-3) log  40/10`

⇒ `t = 1/10^-3 log (4)`

⇒ t = 1000 × 0.602    ...(log (4) = 0.602)

t = 602 seconds = 10 minutes